# On the de Rham cohomology of 1-forms in cotangent bundle.

We know that a cotangent bundle $T^\star M$ has a canonical symplectic form and $M$ is a natural Lagrangian submanifold of it. A well known result is that any submanifold $X=\{(p,f(p)): p\in M\}$, where $f$ is a closed one form is Lagrangian. Denote by $[f]$ the de Rham cohomology class of $f$. Assume that we flow $X$ in a Hamiltonian direction to $Y$, then $Y$ will be a Lagrangian submanifold of $T^\star M$. My question is can we write it as $Y={(p,g(p)): p\in M}$ for some closed one form $g$.? If so, do we have $[g]=[f]?$ Thanks in advance!

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The "well known result" you refer to cannot be true if understood literally: for example, if $Y\subset X$ is a smooth submanifold, then the set of all covectors at points $y\in Y$ that annihilate the tangent subspace $T_yY\subset T_yX$, is a Lagrangian submanifold as well. – Serge Lvovski Jan 9 '13 at 14:24
Tanks serge. I think I made a mistake here. I will edit my question. – Mathboy Jan 9 '13 at 15:37
Should "Hodge homology group" read "de Rham cohomology class"? – Tim Perutz Jan 9 '13 at 17:49
Yes, actually it should be called the de Rham cohomology. Thanks! – Mathboy Jan 10 '13 at 8:39
Did you mean de Rham cohomology clacc of $f$? – Serge Lvovski Jan 10 '13 at 9:31

As has been noted in Peter Michor's answer, a Hamiltonian isotopy can certainly move the graph of a closed one-form to a Lagrangian submanifold that is not the graph of a one-form.

However, in the special case that the starting and ending submanifolds are both graphs of closed one-forms, the de Rham cohomology classes will in fact be equal. (It is important that we are assuming here that the isotopy is Hamiltonian and not just symplectic.) This can be seen as follows:

Let $\lambda=\sum p_idq_i$ be the canonical one-form $T^{\ast}M$, so the symplectic form is $d\lambda$. In general, if $\iota:L\to T^{\ast}M$ is a Lagrangian embedding, then the Lagrangian condition amounts to the statement that $\iota^{\ast}\lambda$ is closed. Consequently, there is a well-defined de Rham cohomology class $[\iota^{\ast}\lambda]\in H^1(L;\mathbb{R})$, generally called the Liouville class. In the special case that $L=M$ and the Lagrangian embedding is a closed one-form $\sigma:M\to T^{\ast}M$ (viewed as a section), the definition of the canonical one-form $\lambda$ is such that $\sigma^{\ast}\lambda=\sigma$. In particular, the Liouville class of a closed one-form is just the cohomology class of the one-form.

I claim now that the Liouville class is invariant under Hamiltonian isotopies of the Lagrangian submanifold. In other words, if $\{\phi_t\}$ is a Hamiltonian isotopy, obtained as the flow of the Hamiltonian vector field $X_H$ of a function $H:T^{\ast}M\to \mathbb{R}$ (i.e. $i_{X_H}d\lambda=-dH$), and if $\iota$ is some Lagrangian embedding, I claim that the Liouville class of $\phi_t\circ \iota$ is independent of $t$. (For ease of notation I'll assume $H$ is time-independent, but the time-dependent case is a straightforward modification.) Indeed this follows fairly quickly from Cartan's magic formula, as follows: $$\frac{d}{dt}(\phi_{t}\circ\iota)^{\ast}\lambda=\iota^{\ast}\left(\frac{d}{dt}\phi_{t}^{\ast}\lambda \right) =\iota^{\ast}\phi_{t}^{\ast}\mathcal{L}_{X_H}\lambda=\iota^{\ast}\phi_{t}^{\ast}(di_{X_H}\lambda+i_{X_H}d\lambda)$$ $$=\iota^{\ast}\phi_{t}^{\ast}d\left(i_{X_H}\lambda-H\right)$$

which is exact; hence the cohomology class of $(\phi_{t}\circ\iota)^{\ast}\lambda$ is indeed independent of $t$. In particular if $\iota$ is equal to the section of $T^{*}M$ given by a closed one-form $\sigma$ and if $\phi_1\circ\iota$ is equal to the section given by $\tau$ then looking at their respective Liouville classes shows that $\sigma$ and $\tau$ are cohomologous.

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Thanks, Dear Mike. This is a perfect answer! – Mathboy Jan 11 '13 at 11:05

You can write $Y$ as the image of 1-form as long as $Y$ meets each fiber of $T^*M\to M$ exactly once and this transversally. For a short time this is so (locally on $M$ if you move the image of a closed one-form by a Hamiltonian flow or even a symplectic flow (this is a smooth curve in the group of symplectic diffeomorphisms). If you look at $M=\mathbb R$, symplectic flows on $T^*M$ are just volume preserving flows on $\mathbb R^2$, and you easily see that one can deform $X$ so that it becomes vertical or a curve meandering wildly, and you can do that faster as you go to $\infty$ on $\mathbb R$, so that at no time $Y$ is still the graph of a 1-form (closed plays no role here, since all 1-forms are closed.

Even if $Y$ stays the graph of a form, the de Rham cohomology class is not constant: Take $M=S^1$ then $T^*M$ is a cylinder, and you can move $X$ just up, which increases the integral, thus the cohomology class.

Edit: Okay, take $T^*\mathbb R$. there every symplectiv flow is Hamiltonian since $H^1=0$. Here the vertical flow is a counterexample.

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Thanks, Peter. This answer is helpful to me. – Mathboy Jan 10 '13 at 12:34
In your example on $S^1$, note that the vertical translation on the cylinder is a symplectic isotopy but is not Hamiltonian (i.e. is not given by the flow of the Hamiltonian vector field of a time-dependent function--its flux is nonzero). So if one is restricting to Hamiltonian flows (as the OP seems to be doing) this example doesn't apply. – Mike Usher Jan 11 '13 at 1:05
Yes, I think if one moves $X$ by a Hamiltonian flow to $Y$, then the integral will be kept and so is the cohomology class. – Mathboy Jan 11 '13 at 8:57